Answer:
multiply the number in each can by the number of cans and add the extras
Step-by-step explanation:
You can add them all up. This is the total of the number in each can and the extras:
8 +8 +8 +8 +8 +8 +8 +8 +8 +8 +8 +8 +8 +8 +8 +7
Or you can take advantage of the invention of multiplication to simplify repeated addition. Multiply the number in each can by the number of cans, then add the extras.
15×8 +7 = 127
\left[x \right] = \left[ 0\right][x]=[0]
Answer:
Sum = 15 8/15, Difference = 1 2/15
Step-by-step explanation:
8⅓, 71/5
A. Sum
8⅓ + 7 1/5
Convert to improper fraction
25/3 + 36/5
Find the LCM of 3 and 5. The result is 15. Divide 15 by the denominator of each fraction and multiply the result obtained with the numerator. The result is shown below:
[(25×5) + (36×3)] / 15
[125 + 108] / 15
233 / 15
Convert to mixed fraction
15 8/15
B. Difference
8⅓ – 7 1/5
Convert to improper fraction
25/3 – 36/5
Find the LCM of 3 and 5. The result is 15. Divide 15 by the denominator of each fraction and multiply the result obtained with the numerator. The result is shown below:
[(25×5) – (36×3)] / 15
[125 – 108] / 15
17 / 15
Convert to mixed fraction
1 2/15
SUMMARY:
Sum = 15 8/15, Difference = 1 2/15
This is a difference of squares problem. You just have to make sure every term is a perfect square and that there is subtraction in the problem.
Step-by-step explanation:
you need to substitute the x in (7x +2y=13) with (x=2y+11) to solve it
